Download - Boundary Scan Architecture

Technical University Tallinn, ESTONIA

Boundary Scan Architecture

TDOinternallogic

T

A

P TDO

TMS

TCK

TDI

BSCTDI

Data_out

Data_in

TDO

TDO

TDI

internallogic

internallogic

internallogic

internallogic

T

A

P

T

A

P

TMS

TCK

T

A

P

TAP


Boundary Scan Architecture

Device ID. Register

Bypass Register

Instruction Register (IR)

TDI

TDO

Bou

ndary

Scan

Registe

rs

Internal logic

Data Register

s


Boundary Scan Cell

From

last cell

Update DR

To next cell

Q

QSET

CLR

D

Clock DR

Test/Normal

1

0

Q

QSET

CLR

D0

1

From system pin

Q

QSET

CLR

D

Q

QSET

CLR

D

Shift DR

To syste

m logic

Used at the input or output pins


Boundary Scan Working Modes

SAMPLE mode:

Get snapshot of normal chip output signals



PRELOAD mode:

Put data on boundary scan chain before next instruction



Extest instruction:

Test off-chip circuits and board-level interconnections



INTEST instruction

Feeds external test patterns in and shifts responses out



Bypass instruction:

Bypasses the corresponding chip using 1-bit register

To TDO

From TDIShift DR

Clock DR Q

QD

SET

CLR



IDCODE instruction:

Connects the component device identification register serially between TDI and TDO in the Shift-DR TAP controller state

Allows board-level test controller or external tester to read out component ID

Required whenever a JEDEC identification register is included in the design

TDOTDI Version Part Number Manufacturer ID 1

4-bitsAny format

16-bitsAny format

11-bitsCoded form of JEDEC


Fault Diagnosis with Boundary Scan

Short

Open

1

0

0

0

0

1

Assume stuck-at-0

Assume wired AND



Short

Open

10

00

00

01

11

Assume stuck-at-0

00

00

00

Assume wired AND

Kautz showed in 1974 that a sufficient condition to detect any pair of short circuited nets was that the “horizontal” codes must be unique for all nets. Therefore the test length is ]log2(N)[



Short

Open

101

000

001

011

110

Assume stuck-at-0

001

001

001

Assume wired AND

All 0-s and all 1-s are forbidden codes because of stuck-at faults Therefore the final test length is ]log2(N+2)[



Short

Open

0 101

0 000

0 001

0 011

1 110

Assume stuck-at-0

1 001

0 001

1 001

Assume wired AND

To improve the diagnostic resolution we have to add one bit more


Synthesis of Testable Circuits

2131 xxxxy

1&

&

x1

x

3x

2

y

x1 x2 x3 y&

&

2131 xxxxy

Test generation:

0 1 1 0 1 0 0 01 0 0 1 0 1 1 01 1 0 0 0 0 1 10 0 1 1 1 1 1 1

4 test patterns are needed



2131 xxxxy

1&

&

x1

x

3x

2

y

&

&

x1 x2 x3

y&

&

Here: Only 3 test patterns are needed

010

010

110

110

110

101

Here: 4 test patterns are needed

Two implementations for the same circuit:

First assignment



32172163151432322310 xxxcxxcxxcxcxxcxcxccy Calculation of constants:

2131 xxxxy

fi x1 x2 x3 y f0 0 0 0 1 1 C0 = f0

f1 0 0 1 0 1 C1 = f0 f1

f2 0 1 0 1 0 C2 = f0 f2

f3 0 1 1 0 0 C3 = f0 f1 f2 f3

f4 1 0 0 0 1 C4 = f0 f4

f5 1 0 1 0 1 C5 = f0 f1 f4 f5

f6 1 1 0 1 1 C6 = f0 f2 f4 f6

f7 1 1 1 1 0 C3 = f0 f1 f2 f3 f4 f5 f6 f7

2131131 xxxxxxy

Given:

New:



Test generation method:

2131131 xxxxxxy

x1 x2 x3

y&

&

011

011

110

110

110

101

x1 x2 x3

0 0 01 1 1

0 1 11 0 11 1 0

&1

&0

0

&1

Roles of test patterns:


Testability as a trade-off

Amusing testability:

Theorem: You can test an arbitrary digital system by only 3 test patterns if you design it approprietly

&011

101001 &

011

101

001

&?

&011

101

001

1010 &011

101001

Solution: System FSM Scan-Path CC NAND

Proof: